我基本上是逐个字符地读取文件，当遇到一个新数字时，我会将其存储为astd::map作为元素等于1的键。越是遇到相同数字的次数越多，我将该键的元素递增：
if(intMap.count(singleCharacter)){
//keyexist just increment count else
//cout<<"exist"<<endl;
int newValue = intMap.at(singleCharacter) + 1; //grabs the value that already exists and we need to incrmenet it by 1 and store the new value in the map.
std::map<int, int>::iterator it = intMap.find(singleCharacter); //finds the single character and increments it by 1.
if (it != intMap.end())
it->second = newValue; //setting the new increment value into the element
}else{
//doesnt exist and and we need to create it and incrmenet by 1
//cout<<"doesnt exist"<<endl;
intMap.insert(pair<int, int>(singleCharacter,1));
cout<<singleCharacter <<" new made : "<<intMap.at(singleCharacter) <<endl;
}
}
for (auto& p : intMap ) {
cout << p.first<<": "<< p.second <<endl;; // "Karl", "George"
}
唯一的问题是，当我尝试打印出地图中的所有值时，它会为我提供随机数，而我不知道它们的来源。
这是我正在读取的文件：
An International Standard Book Number (ISBN) is a code of 10 characters, referred to as ISBN-10,
separated by dashes such as 0-7637-0798-8. An ISBN-10 consists of four parts: a group code, a publisher code,
a code that uniquely identifies the book among those published by a particular publisher, and a check character.
The check character is used to validate an ISBN. For the ISBN 0-7637-0798-8, the group code is 0,
which identifies the book as one from an English-speaking country. The publisher code 7637 is for "Jones and Bartlett Publishers
我得到的输出：
48: 8
49: 3
51: 3
54: 3
55: 8
56: 4
57: 2
我应该得到的输出应该是这样的：
1: and the amount of times it was seen
任何数字都一样。
之所以得到这些“随机”数字，是因为您使用std::map<int,int>而不是std::map<char,int>。您得到的是字符符号的ASCII数字代码及其计数。
因此，您需要更改地图的类型，或将键投射回char：
for (auto& p : intMap ) {
cout << static_cast<char>(p.first) << ": "<< p.second <<endl;
}
注意：std::map::operator[]正是针对这种情况而设计的，在这种情况下，它将使用0初始化缺失值，因此您的所有条件都可以替换为：
intMap[singleCharacter]++;
可以在本文档中找到详细信息。
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